lim x--> 0 + [ (( sin (2x) / f(x) ) - ( x^2 f ( 1/x) ) ]

[peblez]

GIven that the tangent line to the graph of f at ( 0, 0 ) has equation 2y = x and that f has a h orizontal asymptote at infinity with equation y = 2, find the value of

lim x--> 0 + [ (( sin (2x) / f(x) ) - ( x^2 f ( 1/x) ) ]

How do i do this problem, i don't know where to begin

 

[peblez]

Re: Limit Question

Nevermind i got the answer

make the take out the sin2x function by multiplying top and bottom by 2x

so u get 2x/ f( x ) - x^2 f ( 1/x) as lim of x --> 0


if x goes to 0 , f ( 1/ x ) becomes f ( infinity) so that equals 2

so its 2 x / f ( x ) - 2x^2

which becomes 2 x - 2x^2 ( f ( x ) / f ( x )

use l'hopital rule